Here is my design for traversal of a node tree:
struct Leaf1{};
struct Leaf2{};
struct Leaf3{};
struct Leaf4{};
struct Leaf5{};
typedef boost::variant< Leaf4, Leaf5 > Node3;
typedef boost::variant< Leaf2, Leaf3, Node3> Node2;
typedef boost::variant< Node2, Leaf1 > Node1;
class NodeVisitor: public boost::static_visitor<void>
{
public:
template<class Node>
void operator()(const Node& e) const
{
boost::apply_visitor( *this, e );
}
void operator()(const Leaf1& e) const{}
void operator()(const Leaf2& e) const{}
void operator()(const Leaf3& e) const{}
void operator()(const Leaf4& e) const{}
void operator()(const Leaf5& e) const{}
};
So I recursively visit the nodes until I arrive at a leaf. The problem above is that I must add a stub for operater() for each leaf. You can see that I have five such stubs above but have many more in practice. Can you suggest a way of templating this stub?
SOLUTION 1: SFINAE-based technique
This solution is based on the fact that failing to substitute template parameters during the instantiation of a template does not cause a compilation error (Substitution Failure Is Not An Error): instead, that template is simply disregarded for overload resolution. Thus, with some trick you can select which overloads of a certain function template shall be made visible depending on the template arguments provided at instantiation time.
When this technique is used, it is important to make sure that the discriminating conditions which decide the visibility of each overload are mutually exclusive, or ambiguity may arise.
To begin with, you need to define some trait metafunction that helps you determine whether a certain class is a leaf:
// Primary template
template<typename T> struct is_leaf<T> { static const bool value = false; };
// Specializations...
template<> struct is_leaf<Leaf1> { static const bool value = true; };
template<> struct is_leaf<Leaf2> { static const bool value = true; };
...
Then, you can use std::enable_if (or boost::enable_if if you are working with C++98) to select which overload of the call operator should be made visible:
class NodeVisitor: public boost::static_visitor<void>
{
public:
// Based on the fact that boost::variant<> defines a type list called
// "types", but any other way of detecting whether we are dealing with
// a variant is OK
template<typename Node>
typename std::enable_if<
!is_same<typename Node::types, void>::value
>::type
operator()(const Node& e) const
{
boost::apply_visitor( *this, e );
}
// Based on the fact that leaf classes define a static constant value
// called "isLeaf", but any other way of detecting whether we are dealing
// with a leaf is OK
template<typename Leaf>
typename std::enable_if<is_leaf<Leaf>::value>::type
operator()(const Leaf& e) const
{
...
}
};
SOLUTION 2: overload-based technique
If you are working on C++98 and do not want to use boost::enable_if as a replacement for std::enable_if, an alternative approach consists in exploiting overload resolution and an unused argument for discriminating between two overloads of a helper function. First of all, you define two dummy classes:
struct true_type { };
struct false_type { };
Then, you create your is_leaf<> metafunction again, properly specializing it for leaf classes:
// Primary template
template<typename T> struct is_leaf<T> { typedef false_type type; };
// Specializations...
template<> struct is_leaf<Leaf1> { typedef true_type type; };
template<> struct is_leaf<Leaf2> { typedef true_type type; };
...
Finally, you create an instance of one of those dummy types to choose the proper overload of a helper function process():
class NodeVisitor: public boost::static_visitor<void>
{
public:
template<typename T>
void operator()(const T& e) const
{
typedef typename is_leaf<T>::type helper;
process(e, helper());
}
template<typename Node>
void process(const Node& e, false_type) const
{
boost::apply_visitor(*this, e);
}
template<typename Leaf>
void process(const Leaf& e, true_type) const
{
...
}
};
Use an additional level of indirection for the LeafN class, something like:
template <typename LeafType>
struct LeafHolder
{
// has real instance of leaf..
};
Then redefine your variant type
typedef boost::variant< LeafHolder<Leaf4>, LeafHolder<Leaf5> > Node3;
typedef boost::variant< LeafHolder<Leaf2>, LeafHolder<Leaf3>, Node3> Node2;
typedef boost::variant< Node2, LeafHolder<Leaf1 > Node1;
Now your visitor can become:
class NodeVisitor: public boost::static_visitor<void>
{
public:
template<class Node>
void operator()(const Node& e) const
{
boost::apply_visitor( *this, e );
}
// single function to handle all leaves...
template <typename LeafType>
void operator()(const LeafHolder<LeafType>& e) const{}
};
Related
Extremely new to c++ however have a question regarding templates
Suppose I have a simple template class as defined below:
template<typename Collection>
class MySack {
private:
Collection c;
public:
typedef typename Collection::value_type value_type;
void add(const value_type& value) {
c.push_back(value);
}
};
The aim of the class being to accept any type of collection, and allow a user to insert the correct type of value for the specified typename Collection.
The obvious problem is that this is only going to work for types which have a push_back method defined, which means it would work with list however not with set.
I started reading about template specialization to see if that'd be any help, however I don't think this would provide a solution as the type contained within the set would have to be known.
How would this problem be approached in c++?
You can use std::experimental::is_detected and if constexpr to make it work:
template<class C, class V>
using has_push_back_impl = decltype(std::declval<C>().push_back(std::declval<V>()));
template<class C, class V>
constexpr bool has_push_back = std::experimental::is_detected_v<has_push_back_impl, C, V>;
template<typename Collection>
class MySack {
private:
Collection c;
public:
typedef typename Collection::value_type value_type;
void add(const value_type& value) {
if constexpr (has_push_back<Collection, value_type>) {
std::cout << "push_back.\n";
c.push_back(value);
} else {
std::cout << "insert.\n";
c.insert(value);
}
}
};
int main() {
MySack<std::set<int>> f;
f.add(23);
MySack<std::vector<int>> g;
g.add(23);
}
You can switch to insert member function, which has the same syntax for std::vector, std::set, std::list, and other containers:
void add(const value_type& value) {
c.insert(c.end(), value);
}
In C++11, you might also want to create a version for rvalue arguments:
void add(value_type&& value) {
c.insert(c.end(), std::move(value));
}
And, kind-of simulate emplace semantics (not truly in fact):
template <typename... Ts>
void emplace(Ts&&... vs) {
c.insert(c.end(), value_type(std::forward<Ts>(vs)...));
}
...
int main() {
using value_type = std::pair<int, std::string>;
MySack<std::vector<value_type>> v;
v.emplace(1, "first");
MySack<std::set<value_type>> s;
s.emplace(2, "second");
MySack<std::list<value_type>> l;
l.emplace(3, "third");
}
I started reading about template specialization to see if that'd be
any help, however I don't think this would provide a solution as the
type contained within the set would have to be known.
You can partially specialize MySack to work with std::set.
template <class T>
class MySack<std::set<T>> {
//...
};
However, this has the disadvantage that the partial specialization replaces the whole class definition, so you need to define all member variables and functions again.
A more flexible approach is to use policy-based design. Here, you add a template parameter that wraps the container-specific operations. You can provide a default for the most common cases, but users can provide their own policy for other cases.
template <class C, class V = typename C::value_type>
struct ContainerPolicy
{
static void push(C& container, const V& value) {
c.push_back(value);
}
static void pop(C& container) {
c.pop_back();
}
};
template <class C, class P = ContainerPolicy<C>>
class MySack
{
Collection c;
public:
typedef typename Collection::value_type value_type;
void add(const value_type& value) {
P::push(c, value);
}
};
In this case, it is easier to provide a partial template specialization for the default policy, because it contains only the functionality related to the specific container that is used. Other logic can still be captured in the MySack class template without the need for duplicating code.
Now, you can use MySack also with your own or third party containers that do not adhere to the STL style. You simply provide your own policy.
struct MyContainer {
void Add(int value);
//...
};
struct MyPolicy {
static void push(MyContainer& c, int value) {
c.Add(value);
}
};
MySack<MyContainer, MyPolicy> sack;
If you can use at least C++11, I suggest the creation of a template recursive struct
template <std::size_t N>
struct tag : public tag<N-1U>
{ };
template <>
struct tag<0U>
{ };
to manage precedence in case a container can support more than one adding functions.
So you can add, in the private section of your class, the following template helper functions
template <typename D, typename T>
auto addHelper (T && t, tag<2> const &)
-> decltype((void)std::declval<D>().push_back(std::forward<T>(t)))
{ c.push_back(std::forward<T>(t)); }
template <typename D, typename T>
auto addHelper (T && t, tag<1> const &)
-> decltype((void)std::declval<D>().insert(std::forward<T>(t)))
{ c.insert(std::forward<T>(t)); }
template <typename D, typename T>
auto addHelper (T && t, tag<0> const &)
-> decltype((void)std::declval<D>().push_front(std::forward<T>(t)))
{ c.push_front(std::forward<T>(t)); }
Observe that the decltype() part enable they (through SFINAE) only if the corresponding method (push_back(), insert() or push_front()) is enabled.
Now you can write add(), in the public section, as follows
template <typename T>
void add (T && t)
{ addHelper<C>(std::forward<T>(t), tag<2>{}); }
The tag<2> element make so the tag<2> addHelper() method is called, if available (if push_back() is available for type C), otherwise is called the tag<1> method (the insert() one) if available, otherwise the tag<0> method (the push_front() one) is available. Otherwise error.
Also observe the T && t and std::forward<T>(t) part. This way you should select the correct semantic: copy or move.
The following is a full working example
#include <map>
#include <set>
#include <list>
#include <deque>
#include <vector>
#include <iostream>
#include <forward_list>
#include <unordered_map>
#include <unordered_set>
template <std::size_t N>
struct tag : public tag<N-1U>
{ };
template <>
struct tag<0U>
{ };
template <typename C>
class MySack
{
private:
C c;
template <typename D, typename T>
auto addHelper (T && t, tag<2> const &)
-> decltype((void)std::declval<D>().push_back(std::forward<T>(t)))
{ c.push_back(std::forward<T>(t)); }
template <typename D, typename T>
auto addHelper (T && t, tag<1> const &)
-> decltype((void)std::declval<D>().insert(std::forward<T>(t)))
{ c.insert(std::forward<T>(t)); }
template <typename D, typename T>
auto addHelper (T && t, tag<0> const &)
-> decltype((void)std::declval<D>().push_front(std::forward<T>(t)))
{ c.push_front(std::forward<T>(t)); }
public:
template <typename T>
void add (T && t)
{ addHelper<C>(std::forward<T>(t), tag<2>{}); }
};
int main ()
{
MySack<std::vector<int>> ms0;
MySack<std::deque<int>> ms1;
MySack<std::set<int>> ms2;
MySack<std::multiset<int>> ms3;
MySack<std::unordered_set<int>> ms4;
MySack<std::unordered_multiset<int>> ms5;
MySack<std::list<int>> ms6;
MySack<std::forward_list<int>> ms7;
MySack<std::map<int, long>> ms8;
MySack<std::multimap<int, long>> ms9;
MySack<std::unordered_map<int, long>> msA;
MySack<std::unordered_multimap<int, long>> msB;
ms0.add(0);
ms1.add(0);
ms2.add(0);
ms3.add(0);
ms4.add(0);
ms5.add(0);
ms6.add(0);
ms7.add(0);
ms8.add(std::make_pair(0, 0L));
ms9.add(std::make_pair(0, 0L));
msA.add(std::make_pair(0, 0L));
msB.add(std::make_pair(0, 0L));
}
To implement a property system for polymorphic objects, I first declared the following structure:
enum class access_rights_t
{
NONE = 0,
READ = 1 << 0,
WRITE = 1 << 1,
READ_WRITE = READ | WRITE
};
struct property_format
{
type_index type;
string name;
access_rights_t access_rights;
};
So a property is defined with a type, a name and access rights (read-only, write-only or read-write). Then I started the property class as follows:
template<typename Base>
class property : property_format
{
public:
template<typename Derived, typename T>
using get_t = function<T(const Derived&)>;
template<typename Derived, typename T>
using set_t = function<void(Derived&, const T&)>;
private:
get_t<Base, any> get_f;
set_t<Base, any> set_f;
The property is associated to a base type, but may (and will) be filled with accessors associated to an instance of a derived type. The accessors will be encapsulated with functions accessing std::any objects on an instance of type Base. The get and set methods are declared as follows (type checking are not shown here to make the code minimal):
public:
template<typename T>
T get(const Base& b) const
{
return any_cast<T>(this->get_f(b));
}
template<typename T>
void set(Base& b, const T& value_)
{
this->set_f(b, any(value_));
}
Then the constructors (access rights are set to NONE to make the code minimal):
template<typename Derived, typename T>
property(
const string& name_,
get_t<Derived, T> get_,
set_t<Derived, T> set_ = nullptr
):
property_format{
typeid(T),
name_,
access_rights_t::NONE
},
get_f{caller<Derived, T>{get_}},
set_f{caller<Derived, T>{set_}}
{
}
template<typename Derived, typename T>
property(
const string& name_,
set_t<Derived, T> set_
):
property{
name_,
nullptr,
set_
}
{
}
The functions passed as arguments are encapsulated through the helper structure caller:
private:
template<typename Derived, typename T>
struct caller
{
get_t<Derived, T> get_f;
set_t<Derived, T> set_f;
caller(get_t<Derived, T> get_):
get_f{get_}
{
}
caller(set_t<Derived, T> set_):
set_f{set_}
{
}
any operator()(const Base& object_)
{
return any{
this->get_f(
static_cast<const Derived&>(object_)
)
};
}
void operator()(Base& object_, const any& value_)
{
this->set_f(
static_cast<Derived&>(object_),
any_cast<Value>(value_)
);
}
};
Now, considering these dummy classes.
struct foo
{
};
struct bar : foo
{
int i, j;
bar(int i_, int j_):
i{i_},
j{j_}
{
}
int get_i() const {return i;}
void set_i(const int& i_) { this->i = i_; }
};
I can write the following code:
int main()
{
// declare accessors through bar methods
property<foo>::get_t<bar, int> get_i = &bar::get_i;
property<foo>::set_t<bar, int> set_i = &bar::set_i;
// declare a read-write property
property<foo> p_i{"bar_i", get_i, set_i};
// declare a getter through a lambda
property<foo>::get_t<bar, int> get_j = [](const bar& b_){ return b_.j; };
// declare a read-only property
property<foo> p_j{"bar_j", get_j};
// dummy usage
bar b{42, 24};
foo& f = b;
cout << p_i.get<int>(f) << " " << p_j.get<int>(f) << endl;
p_i.set<int>(f, 43);
cout << p_i.get<int>(f) << endl;
}
My problem is that template type deduction doesn't allow me to declare a property directly passing the accessors as arguments, as in:
property<foo> p_i{"bar_i", &bar::get_i, &bar::set_i};
Which produces the following error:
prog.cc:62:5: note: template argument deduction/substitution failed:
prog.cc:149:50: note: mismatched types std::function<void(Type&, const Value&)> and int (bar::*)() const
property<foo> p_i{"bar_i", &bar::get_i, set_i};
Is there a way to address this problem while keeping the code "simple"?
A complete live example is available here.
std::function is a type erasure type. Type erasure types are not suitable for deduction.
template<typename Derived, typename T>
using get_t = function<T(const Derived&)>;
get_t is an alias to a type erasure type. Ditto.
Create traits classes:
template<class T>
struct gettor_traits : std::false_type {};
this will tell you if T is a valid gettor, and if so what its input and output types are. Similarly for settor_traits.
So
template<class T, class Derived>
struct gettor_traits< std::function<T(Derived const&)> >:
std::true_type
{
using return_type = T;
using argument_type = Derived;
};
template<class T, class Derived>
struct gettor_traits< T(Derived::*)() >:
std::true_type
{
using return_type = T;
using argument_type = Derived;
};
etc.
Now we got back to the property ctor:
template<class Gettor,
std::enable_if_t< gettor_traits<Gettor>{}, int> =0,
class T = typename gettor_traits<Gettor>::return_value,
class Derived = typename gettor_traits<Gettor>::argument_type
>
property(
const string& name_,
Gettor get_
):
property_format{
typeid(T),
name_,
access_rights_t::NONE
},
get_f{caller<Derived, T>{get_}},
nullptr
{
}
where we use SFINAE to ensure that our Gettor passes muster, and the traits class to extract the types we care about.
There is going to be lots of work here. But it is write-once work.
My preferred syntax in these cases would be:
std::cout << (f->*p_i)();
and
(f->*p_i)(7);
where the property acts like a member function pointer, or even
(f->*p_i) = 7;
std::cout << (f->*p_i);
where the property transparently acts like a member variable pointer.
In both cases, through overload of ->*, and in the second case via returning a pseudo-reference from ->*.
At the end of this answer is a slightly different approach. I will begin with the general problem though.
The problem is &bar::get_i is a function pointer to a member function while your alias is creating a function object which needs the class as additional template parameter.
Some examples:
Non member function:
#include <functional>
void a(int i) {};
void f(std::function<void(int)> func)
{
}
int main()
{
f(&a);
return 0;
}
This works fine. Now if I change a into a struct:
#include <functional>
struct A
{
void a(int i) {};
};
void f(std::function<void(int)> func)
{
}
int main()
{
f(std::function<void(int)>(&A::a));
return 0;
}
this gets the error:
error: no matching function for call to std::function<void(int)>::function(void (A::*)(int))'
because the std::function object also need the base class (as you do with your alias declaration)
You need a std::function<void(A,int)>
You cannot make your example much better though.
A way to make it a "bit" more easier than your example would maybe be this approach using CRTP.
#include <functional>
template <typename Class>
struct funcPtr
{
template <typename type>
using fun = std::function<void(Class,type)>;
};
struct A : public funcPtr<A>
{
void a(int i) {};
};
void f(A::fun<int> func)
{
};
int main()
{
f(A::fun<int>(&A::a));
return 0;
}
And each your "derived" classes derives from a funcPtr class which "auto generates" the specific alias declaration.
I have a typical type-erasure setup:
struct TEBase
{
virtual ~TEBase() {}
// ...
};
template <typename T>
struct TEImpl : TEBase
{
// ...
};
Now the question: Given a second class hierarchy like this,
struct Foo { };
struct Bar : Foo { };
struct Unrelated { };
is it possible, given a TEBase * p, to determine whether the dynamic type of *p is of the form TEImpl<X>, where, X derives from Foo? In other words, I want function:
template <typename T> bool is_derived_from(TEBase * p);
such that:
is_derived_from<Foo>(new TEImpl<Foo>) == true
is_derived_from<Foo>(new TEImpl<Bar>) == true
is_derived_from<Foo>(new TEImpl<Unrelated>) == false
In particular, I'm looking for a solution that is general, non-intrusive, and efficient. I've found two solutions to this problem (posted below as answers) but neither of them solve all three criteria.
Something like this:
template <typename Type, typename UnaryPredicate>
void DoPred(UnaryPredicate pred)
{
if (T * p = dynamic_cast<Derived<T> *>(this))
{
return pred(p->type);
}
return false;
}
This isn't 100% universal, since you cannot, for example, say DoPred<int>. A more universal solution would add a virtual std::type_info type() const { return typeid(...); } member function to the hierarchy and use that to determine if the type matches (the standard type erasure idiom). Both approaches use the same sort of RTTI, though.
After the clarification:
Right now, I don't think this can be solved. All you have is a TEBase subobject. It could be part of a TEImpl<Bar>, or part of a TEImpl<Unrelated>, but neither of those types is related to TEImpl<Foo>, which is what you're after.
You're essentially asking that TEImpl<Bar> derives from TEImpl<Foo>. To do this, you would actually want TEImpl<T> to inherit from all TEImpl<std::direct_bases<T>::type>..., if you see what I mean. This is not possible in C++11, but will be possible in TR2. GCC already supports it. Here is an example implementation. (It causes a warning due to ambiguous bases, which could be avoided with more work, but it works nonetheless.)
#include <tr2/type_traits>
struct TEBase { virtual ~TEBase() {} };
template <typename T> struct TEImpl;
template <typename TL> struct Derivator;
template <typename TL, bool EmptyTL>
struct DerivatorImpl;
template <typename TL>
struct DerivatorImpl<TL, true>
: TEBase
{ };
template <typename TL>
struct DerivatorImpl<TL, false>
: TEImpl<typename TL::first::type>
, Derivator<typename TL::rest::type>
{ };
template <typename TL>
struct Derivator
: DerivatorImpl<TL, TL::empty::value>
{ };
template <typename T>
struct TEImpl
: Derivator<typename std::tr2::direct_bases<T>::type>
{
};
template <typename T>
bool is(TEBase const * b)
{
return nullptr != dynamic_cast<TEImpl<T> const *>(b);
}
struct Foo {};
struct Bar : Foo {};
struct Unrelated {};
#include <iostream>
#include <iomanip>
int main()
{
TEImpl<int> x;
TEImpl<Unrelated> y;
TEImpl<Bar> z;
TEImpl<Foo> c;
std::cout << std::boolalpha << "int ?< Foo: " << is<Foo>(&x) << "\n";
std::cout << std::boolalpha << "Unr ?< Foo: " << is<Foo>(&y) << "\n";
std::cout << std::boolalpha << "Bar ?< Foo: " << is<Foo>(&z) << "\n";
std::cout << std::boolalpha << "Foo ?< Foo: " << is<Foo>(&c) << "\n";
}
I would suggest reading the article Generic Programming:Typelists and Applications. There Andrei Alexandrescu desribes an implementation of a ad-hoc Visitor which should solve your problem. Another good resource would be his book Moder C++ Design where he describes a multidispatcher in a Brute Force way which uses the same approuch (pages 265 ...).
In my opinion these 2 resources are better for understanding than any code which could be printed here.
This solution involves abusing exceptions a bit. If the TEImpl type simply throws its data, is_derived_from can catch the type it's looking for.
struct TEBase
{
virtual ~TEBase() {}
virtual void throw_data() = 0;
};
template <typename T>
struct TEImpl : public TEBase
{
void throw_data() {
throw &data;
}
T data;
};
template <typename T>
bool is_derived_from(TEBase* p)
{
try {
p->throw_data();
} catch (T*) {
return true;
} catch (...) {
// Do nothing
}
return false;
}
This solution works great. It works perfectly with any inheritance structure, and it's completely non-intrusive.
The only problem is that it's no efficient at all. Exceptions were not intended to be used in this way, and I suspect this solution is thousands of times slower than other solutions.
This solution involves comparing typeids. TEImpl knows its own type, so it can check a passed typeid against its own.
The trouble is, this technique doesn't work when you add inheritance, so I'm also using template meta-programming to check if the type has typedef super defined, in which case it will recursively check its parent class.
struct TEBase
{
virtual ~TEBase() {}
virtual bool is_type(const type_info& ti) = 0;
};
template <typename T>
struct TEImpl : public TEBase
{
bool is_type(const type_info& ti) {
return is_type_impl<T>(ti);
}
template <typename Haystack>
static bool is_type_impl(const type_info& ti) {
return is_type_super<Haystack>(ti, nullptr);
}
template <typename Haystack>
static bool is_type_super(const type_info& ti, typename Haystack::super*) {
if(typeid( Haystack ) == ti) return true;
return is_type_impl<typename Haystack::super>(ti);
}
template <typename Haystack>
static bool is_type_super(const type_info& ti, ...) {
return typeid(Haystack) == ti;
}
};
template <typename T>
bool is_derived_from(TEBase* p)
{
return p->is_type(typeid( T ));
}
For this to work with, Bar needs to be redefined as:
struct Bar : public Foo
{
typedef Foo super;
};
This should be fairly efficient, but it's obviously not non-intrusive, since it requires a typedef super in the target class whenever inheritance is being used. The typedef super also has to be publicly accessible, which goes against what many consider to be a recommended practice of putting your typedef super in your private section.
It also doesn't deal with multiple-inheritance at all.
Update: This solution can be taken further to make it general and non-intrusive.
Making it general
typedef super is great, because it's idiomatic and already used in many classes, but it doesn't allow multiple inheritance. In order to do that, we'll need to replace it with a type that can store multiple types, such as a tuple.
If Bar was rewritten as:
struct Bar : public Foo, public Baz
{
typedef tuple<Foo, Baz> supers;
};
we could support this form of declaration by adding the following code to TEImpl:
template <typename Haystack>
static bool is_type_impl(const type_info& ti) {
// Redefined to call is_type_supers instead of is_type_super
return is_type_supers<Haystack>(ti, nullptr);
}
template <typename Haystack>
static bool is_type_supers(const type_info& ti, typename Haystack::supers*) {
return IsTypeTuple<typename Haystack::supers, tuple_size<typename Haystack::supers>::value>::match(ti);
}
template <typename Haystack>
static bool is_type_supers(const type_info& ti, ...) {
return is_type_super<Haystack>(ti, nullptr);
}
template <typename Haystack, size_t N>
struct IsTypeTuple
{
static bool match(const type_info& ti) {
if(is_type_impl<typename tuple_element< N-1, Haystack >::type>( ti )) return true;
return IsTypeTuple<Haystack, N-1>::match(ti);
}
};
template <typename Haystack>
struct IsTypeTuple<Haystack, 0>
{
static bool match(const type_info& ti) { return false; }
};
Making it non-intrusive
Now we have a solution which is efficient and general, but it's still intrusive, so it won't support classes that can't be modified.
To support this, we'll need a way to declare the object inheritance from outside the class. For Foo, we could do something like this:
template <>
struct ClassHierarchy<Bar>
{
typedef tuple<Foo, Baz> supers;
};
To support that style, first we need the non-specialized form of ClassHierarchy, which we'll define like so:
template <typename T> struct ClassHierarchy { typedef bool undefined; };
We'll use the presence of undefined to tell whether or not the class has been specialized.
Now we need to add some more functions to TEImpl. We'll still reuse most of the code from earlier, but now we'll also support reading the type data from ClassHierarchy.
template <typename Haystack>
static bool is_type_impl(const type_info& ti) {
// Redefined to call is_type_external instead of is_type_supers.
return is_type_external<Haystack>(ti, nullptr);
}
template <typename Haystack>
static bool is_type_external(const type_info& ti, typename ClassHierarchy<Haystack>::undefined*) {
return is_type_supers<Haystack>(ti, nullptr);
}
template <typename Haystack>
static bool is_type_external(const type_info& ti, ...) {
return is_type_supers<ClassHierarchy< Haystack >>(ti, nullptr);
}
template <typename Haystack>
struct ActualType
{
typedef Haystack type;
};
template <typename Haystack>
struct ActualType<ClassHierarchy< Haystack >>
{
typedef Haystack type;
};
template <typename Haystack>
static bool is_type_super(const type_info& ti, ...) {
// Redefined to reference ActualType
return typeid(typename ActualType<Haystack>::type) == ti;
}
And now we have a solution which is efficient, general, and non-intrusive.
Future solution
This solution meets the criteria, but it's still a little annoying to have to document the class hierarchy explicitly. The compiler already knows everything about the class hierarchy, so it's a shame that we have to do this grunt work.
A proposed solution to this problem is N2965: Type traits and base classes, which has been implemented in GCC. This paper defines a direct_bases class, which is almost identical to our ClassHierarchy class, except its only element, type, is guaranteed to be a tuple, like supers, and the class is completely generated by the compiler.
So for now we have to write a little boilerplate to get this to work, but if N2965 gets accepted, we can get rid of the boilerplate and make TEImpl much shorter.
Special thanks to Kerrek SB and Jan Herrmann. This answer drew a lot of inspiration from their comments.
I want to automatically choose the right pointer-to-member among overloaded ones based on the "type" of the member, by removing specializations that accept unconcerned members (via enable_if).
I have the following code:
class test;
enum Type
{
INT_1,
FLOAT_1,
UINT_1,
CHAR_1,
BOOL_1,
INT_2,
FLOAT_2,
UINT_2,
CHAR_2,
BOOL_2
};
template<typename T, Type Et, typename func> struct SetterOk { static const bool value = false; };
template<typename T> struct SetterOk<T,INT_1,void (T::*)(int)> { static const bool value = true; };
template<typename T> struct SetterOk<T,FLOAT_1,void (T::*)(float)> { static const bool value = true; };
template<typename T> struct SetterOk<T,UINT_1,void (T::*)(unsigned int)> { static const bool value = true; };
template<typename T> struct SetterOk<T,CHAR_1,void (T::*)(char)> { static const bool value = true; };
template<typename T> struct SetterOk<T,BOOL_1,void (T::*)(bool)> { static const bool value = true; };
template<typename T> struct SetterOk<T,INT_2,void (T::*)(int,int)> { static const bool value = true; };
template<typename T> struct SetterOk<T,FLOAT_2,void (T::*)(float,float)> { static const bool value = true; };
template<typename T> struct SetterOk<T,UINT_2,void (T::*)(unsigned int, unsigned int)> { static const bool value = true; };
template<typename T> struct SetterOk<T,CHAR_2,void (T::*)(char,char)> { static const bool value = true; };
template<typename T> struct SetterOk<T,BOOL_2,void (T::*)(bool,bool)> { static const bool value = true; };
template <bool, class T = void> struct enable_if {};
template <class T> struct enable_if<true, T> { typedef T type; };
template<typename T, Type Et>
struct Helper
{
template<typename U>
static void func(U method, typename enable_if<SetterOk<T,Et,U>::value>::type* dummy = 0)
{
}
};
class test
{
public:
void init()
{
Helper<test,INT_2>::func(&test::set);
}
void set2(int);
void set(int);
void set(int,int);
void set(float,float);
};
int main()
{
test t;
t.init();
return 0;
}
I'm expecting it to choose the right function between all possible. The problem is that the compiler says "cannot deduce template argument as function argument is ambiguous".
It seems I don't know how to use enable_if, because if so the compiler would only allow the specialization if the specified function has the right type...
Note that I want to have C++03 solutions (if possible) - my code must compile on some old compilers.
Thanks in advance
You can never refer to an overloaded function without disambiguating it (means: static_casting it to the correct type). When you instantiate Helper::func the type of the function argument cannot be known without ever disambiguating it.
The reason it doesn't compile is quite simply that there are several different overloaded functions and it doesn't know which one you mean. Granted, only one of these (void set(int,int)) would actually compile, given the specialization Helper<test,INT_2>. However, this is not enough for the compiler to go on.
One way of getting this to compile would be to explicitly cast &test::set to the appropriate type:
Helper<test,INT_2>::func(static_cast<void (test::*)(int,int)>(&test::set));
Another way would be to use explicit template specialization:
Helper<test,INT_2>::func<void (test::*)(int,int)>((&test::set));
Either way, you need to let the compiler know which of the set functions you are trying to refer to.
EDIT:
As I understand it, you want to be able to deduce, from the use of a Type, which function type should be used. The following alternative achieves this:
template<typename T, Type Et> struct SetterOK{};
template<typename T> struct SetterOK<T,INT_1> {typedef void (T::*setter_type)(int);};
template<typename T> struct SetterOK<T,FLOAT_1> {typedef void (T::*setter_type) (float);};
// ...
template<typename T> struct SetterOK<T,INT_2> {typedef void (T::*setter_type)(int,int);};
// ....
template<typename T, Type Et>
struct Helper
{
template<typename U>
static void func(U method)
{
}
};
class test
{
public:
void init()
{
Helper<test,INT_2>::func<SetterOK<test,INT_2>::setter_type >(&test::set);
}
void set2(int);
void set(int);
void set(int,int);
void set(float,float);
};
int main()
{
test t;
t.init();
return 0;
}
ADDITIONAL EDIT:
A thought just occurred to me. In this special case which you've done, where U is SetterOK::setter_type, things can be simplified further by completely removing the template arguments for func:
static void func(typename SetterOK<T,Et>::setter_type method)
{
}
This would make the init method a simpler:
void init()
{
Helper<test,INT_2>::func(&test::set);
}
I have a sfinae class that tests whether a class is a parser rule (AXE parser generator library).
The axe::is_rule<P>::value should evaluate to true iff P satisfies parser rule requirements. A parser rule must have one of the following member functions, taking a pair of iterators and returning axe::result<Iterator>:
template<class Iterator>
axe::result<Iterator> P::operator()(Iterator, Iterator);
, or its specialization, or non-template for some type CharT
axe::result<CharT*> P::operator()(CharT*, CharT*);
, or const versions of the above. Theoretically, there can be more than one overloaded operator(), though in practice a test for a single operator() with one of the above signatures would suffice.
Unfortunately, current implementation of is_rule takes care of only some, but not all cases. There are some unfortunate classes, that fail the is_rule test:
#define AXE_ASSERT_RULE(T)\
static_assert(axe::is_rule<typename std::remove_reference<T>::type>::value, \
"type '" #T "' is not a rule");
For example, the following unfortunate types fail the test:
struct unfortunate
{
axe::result<const unsigned char*>
operator()(const unsigned char*, const unsigned char*);
};
AXE_ASSERT_RULE(unfortunate);
// or same using lambda
auto unfortunate1 = [](const unsigned char*, const unsigned char*)
->axe::result<const unsigned char*> {};
AXE_ASSERT_RULE(decltype(unfortunate1));
typedef std::vector<char>::iterator vc_it;
struct unfortunate2 { axe::result<vc_it> operator()(vc_it, vc_it) const; };
AXE_ASSERT_RULE(unfortunate2);
typedef axe::result<const char*> (unfortunate3)(const char*, const char*);
AXE_ASSERT_RULE(unfortunate3);
struct rule { template<class I> axe::result<I> operator()(I, I); };
class unfortunate4 : public rule {};
AXE_ASSERT_RULE(unfortunate4);
Current solution in AXE is to wrap those in a forwarding wrapper (class r_ref_t), which, of course, creates syntactic warts (after all, parser generator is all about syntactic sugar).
How would you modify the sfinae test in is_rule to cover the unfortunate cases above?
I think the API of is_rule is not sufficient. For example unfortunate is a rule only if used with iterators of type const unsigned char*. If you use unfortunate with const char*, then it doesn't work, and is thus not a rule, right?
That being said, if you change the API to:
template <class R, class It> struct is_rule;
then I think this is doable in C++11. Below is a prototype:
#include <type_traits>
namespace axe
{
template <class It>
struct result
{
};
}
namespace detail
{
struct nat
{
nat() = delete;
nat(const nat&) = delete;
nat& operator=(const nat&) = delete;
~nat() = delete;
};
struct any
{
any(...);
nat operator()(any, any) const;
};
template <class T>
struct wrap
: public any,
public T
{
};
template <bool, class R, class It>
struct is_rule
{
typedef typename std::conditional<std::is_const<R>::value,
const wrap<R>,
wrap<R>>::type W;
typedef decltype(
std::declval<W>()(std::declval<It>(), std::declval<It>())
) type;
static const bool value = std::is_convertible<type, axe::result<It>>::value;
};
template <class R, class It>
struct is_rule<false, R, It>
{
static const bool value = false;
};
} // detail
template <class R, class It>
struct is_rule
: public std::integral_constant<bool,
detail::is_rule<std::is_class<R>::value, R, It>::value>
{
};
struct unfortunate
{
axe::result<const unsigned char*>
operator()(const unsigned char*, const unsigned char*);
};
#include <iostream>
int main()
{
std::cout << is_rule<unfortunate, const unsigned char*>::value << '\n';
std::cout << is_rule<unfortunate, const char*>::value << '\n';
}
For me this prints out:
1
0
I made the rule slightly more lax than you specified: The return type only has to be implicitly convertible to axe::result<It>. If you really want it to be exactly axe::result<It> then just sub in std::is_same where I used std::is_convertible.
I also made is_rule derive from std::integral_constant. This can be very convenient for tag dispatching. E.g.:
template <class T>
void imp(T, std::false_type);
template <class T>
void imp(T, std::true_type);
template <class T>
void foo(T t) {imp(t, is_rule<T, const char*>());}